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- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
- 4x^2+y+20x-11=0
-
xy+x+y=0
- 3*x^2-6*x*y+3*z^2+6*x*z-6*y*z-2*x-6*y-z-4=0
- 25x^2+16y^2+10x+8y+36=0
- Plot:
- 2*x^2+6*x+9
- x^2-2=y
-
48x²y²+9+64x⁴y⁴
- 3*x-84*y=54
-
Identical expressions
- two 5x^ two +16y^2+1 zero x+8y+ thirty-six =0
- 25x squared plus 16y squared plus 10x plus 8y plus 36 equally 0
- two 5x to the power of two plus 16y squared plus 1 zero x plus 8y plus thirty minus six equally 0
- 25x2+16y2+10x+8y+36=0
- 25x²+16y²+10x+8y+36=0
- 25x to the power of 2+16y to the power of 2+10x+8y+36=0
- 25x^2+16y^2+10x+8y+36=O
-
Similar expressions
- 25x^2+16y^2-10x+8y+36=0
- 25x^2+16y^2+10x+8y-36=0
- 25x^2+16y^2+10x-8y+36=0
- 25x^2-16y^2+10x+8y+36=0
25x^2+16y^2+10x+8y+36=0 canonical form
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The solution
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2 2 36 + 8*y + 10*x + 16*y + 25*x = 0
$$25 x^{2} + 10 x + 16 y^{2} + 8 y + 36 = 0$$
25*x^2 + 10*x + 16*y^2 + 8*y + 36 = 0
Detail solution
Given line equation of 2-order:
$$25 x^{2} + 10 x + 16 y^{2} + 8 y + 36 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 5$$
$$a_{22} = 16$$
$$a_{23} = 4$$
$$a_{33} = 36$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & 0\\0 & 16\end{matrix}\right|$$
$$\Delta = 400$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$25 x_{0} + 5 = 0$$
$$16 y_{0} + 4 = 0$$
then
$$x_{0} = - \frac{1}{5}$$
$$y_{0} = - \frac{1}{4}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = 5 x_{0} + 4 y_{0} + 36$$
$$a'_{33} = 34$$
then equation turns into
$$25 x'^{2} + 16 y'^{2} + 34 = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{5 \frac{\sqrt{34}}{34}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{4 \frac{\sqrt{34}}{34}}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
$$25 x^{2} + 10 x + 16 y^{2} + 8 y + 36 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 5$$
$$a_{22} = 16$$
$$a_{23} = 4$$
$$a_{33} = 36$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}25 & 0\\0 & 16\end{matrix}\right|$$
$$\Delta = 400$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$25 x_{0} + 5 = 0$$
$$16 y_{0} + 4 = 0$$
then
$$x_{0} = - \frac{1}{5}$$
$$y_{0} = - \frac{1}{4}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = 5 x_{0} + 4 y_{0} + 36$$
$$a'_{33} = 34$$
then equation turns into
$$25 x'^{2} + 16 y'^{2} + 34 = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{5 \frac{\sqrt{34}}{34}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{4 \frac{\sqrt{34}}{34}}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
(-1/5, -1/4)
Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$25 x^{2} + 10 x + 16 y^{2} + 8 y + 36 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 5$$
$$a_{22} = 16$$
$$a_{23} = 4$$
$$a_{33} = 36$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 41$$
$$I_{3} = \left|\begin{matrix}25 & 0 & 5\\0 & 16 & 4\\5 & 4 & 36\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & 0\\0 & 16 - \lambda\end{matrix}\right|$$
$$I_{1} = 41$$
$$I_{2} = 400$$
$$I_{3} = 13600$$
$$I{\left(\lambda \right)} = \lambda^{2} - 41 \lambda + 400$$
$$K_{2} = 1435$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} > 0$$
then by line type:
this equation is of type : imaginary ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 41 \lambda + 400 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = 16$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$25 \tilde x^{2} + 16 \tilde y^{2} + 34 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{5 \frac{\sqrt{34}}{34}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{4 \frac{\sqrt{34}}{34}}\right)^{2}} = -1$$
- reduced to canonical form
$$25 x^{2} + 10 x + 16 y^{2} + 8 y + 36 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 5$$
$$a_{22} = 16$$
$$a_{23} = 4$$
$$a_{33} = 36$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
|a11 a12|
I2 = | |
|a12 a22|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
|a11 a13| |a22 a23|
K2 = | | + | |
|a13 a33| |a23 a33|substitute coefficients
$$I_{1} = 41$$
|25 0 |
I2 = | |
|0 16|$$I_{3} = \left|\begin{matrix}25 & 0 & 5\\0 & 16 & 4\\5 & 4 & 36\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & 0\\0 & 16 - \lambda\end{matrix}\right|$$
|25 5 | |16 4 |
K2 = | | + | |
|5 36| |4 36|$$I_{1} = 41$$
$$I_{2} = 400$$
$$I_{3} = 13600$$
$$I{\left(\lambda \right)} = \lambda^{2} - 41 \lambda + 400$$
$$K_{2} = 1435$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} > 0$$
then by line type:
this equation is of type : imaginary ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
or
$$\lambda^{2} - 41 \lambda + 400 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = 16$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$25 \tilde x^{2} + 16 \tilde y^{2} + 34 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{5 \frac{\sqrt{34}}{34}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{4 \frac{\sqrt{34}}{34}}\right)^{2}} = -1$$
- reduced to canonical form